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# A certain car traveled twice as many miles from Town A

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Intern
Joined: 17 Jan 2012
Posts: 41
GMAT 1: 610 Q43 V31
A certain car traveled twice as many miles from Town A  [#permalink]

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21 Feb 2012, 14:58
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10
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65% (hard)

Question Stats:

62% (01:45) correct 38% (01:26) wrong based on 314 sessions

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A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

A. 13
B. 13.5
C. 14
D. 14.5
E. 15

MGMAT 1 Q. 11
The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!
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Re: A certain car traveled twice as many miles from Town A  [#permalink]

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21 Feb 2012, 21:58
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3
docabuzar wrote:
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

A. 13
B. 13.5
C. 14
D. 14.5
E. 15

MGMAT 1 Q. 11
The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!

If the concept you do not understand is why (12*2 + 18*1)/3 doesn't work, here you go:

'Distances traveled' (i.e. ratio of 2:1) cannot be the weights here to find the average mileage. The weights have to be 'number of gallons'.

If we change the question and make it:
A certain car used [highlight]twice as many gallons[/highlight] from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

Now you can use (12*2 + 18 *1)/3

Why?

Whenever you are confused what the weights should be (e.g. here should the weights be the distance traveled or should they be gallons of fuel used...), look at the units.

Average required is $$\frac{miles}{gallon}$$. So you are trying to find the weighted average of two quantities whose units must be $$\frac{miles}{gallon}$$.

$$C_{avg} = \frac{C_1*W_1 + C_2*W_2}{{W_1 + W_2}}$$

$$C_{avg}, C_1, C_2 - \frac{miles}{gallon}$$

So $$W_1$$ and $$W_2$$ should be in gallon to get:

$$\frac{miles}{gallon} = (\frac{miles}{gallon}*gallon + \frac{miles}{gallon}*gallon)/(gallon + gallon)$$

Only if weights are in gallons, do we get 'Total miles' in the numerator and 'Total gallons' in the denominator.

We know that Average miles/gallon = Total miles/Total gallons

[highlight]Takeaway: The weights have to be the denominator units of the average.[/highlight]

So what do we do in this question?

What is the distance travelled (i.e.total miles)?
"traveled twice as many miles from Town A to Town B as it did from Town B to Town C"
We know that the ratio of the two distances must be 2:1 or we can say the distances must be 2d and d. Total distance must be (2d + d)

What is the total gallons used?
Fuel used to go from A to B = 2d/12
Fuel used to go from B to C = d/18

So Average miles/gallon = (2d + d)/(2d/12 + d/18)

You need to know what the weights are going to be. Here, weights have to be the amount of fuel used i.e. in gallons because you are looking for average miles per gallon.
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##### General Discussion
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Joined: 02 Sep 2009
Posts: 49272
Re: A certain car traveled twice as many miles from Town A  [#permalink]

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21 Feb 2012, 15:15
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4
docabuzar wrote:
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

A. 13
B. 13.5
C. 14
D. 14.5
E. 15

MGMAT 1 Q. 11
The explanation given is unclear to me, why weightage ave is not right to apply here? I want to understand conceptually. Can someone help!

Average miles per gallon equals to total miles covered/total gallons used (miles/gallons), so if the distance from A to B is $$2x$$ miles and from B to C is $$x$$ miles then we'll have:
$$average \ miles \ per \ gallon=\frac{total \ miles \ covered}{total \ gallons \ used}=\frac{2x+x}{\frac{2x}{12}+\frac{x}{18}}=\frac{3x}{\frac{2x}{9}}=13.5$$.

P.S. As you can see weighted average concept is exactly what we used here.

Hope it's clear.
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Re: A certain car traveled twice as many miles from Town A  [#permalink]

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21 Feb 2012, 15:40
3
Lets pick numbers:

Distance A to B: 36 miles (2x) --> 3 gallons
Distance B to C: 18 miles (x) --> 1 gallon

Average: $$\frac{(36+18)}{4}$$ = 13.5

Hence, B
Intern
Joined: 17 Jan 2012
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Re: A certain car traveled twice as many miles from Town A  [#permalink]

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22 Feb 2012, 04:10
Many Thanks.

Just to make myself clearer, we can also say that:

Case 1.
Car moves A to B @ 12mph and covers 2 miles
B to C @ 18 mph and covers 1 mile

We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph
because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing).

Case 2.
Car moves A to B @ 12mph for 2 hrs
B to C @ 18 mph for 1 hr

We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph

But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here!
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Joined: 02 Sep 2009
Posts: 49272
Re: A certain car traveled twice as many miles from Town A  [#permalink]

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22 Feb 2012, 04:50
docabuzar wrote:
Many Thanks.

Just to make myself clearer, we can also say that:

Case 1.
Car moves A to B @ 12mph and covers 2 miles
B to C @ 18 mph and covers 1 mile

We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph
because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing).

Case 2.
Car moves A to B @ 12mph for 2 hrs
B to C @ 18 mph for 1 hr

We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph

But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here!

I think that the issue was over complicated above:
{average rate}={total distance}/{total time};
{average salary}={total salary}/{total # of employees};
{average miles per gallon}={total miles}/{total gallons};
...

Keep it simple.
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Re: A certain car traveled twice as many miles from Town A  [#permalink]

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22 Feb 2012, 05:30
docabuzar wrote:
Many Thanks.

Just to make myself clearer, we can also say that:

Case 1.
Car moves A to B @ 12mph and covers 2 miles
B to C @ 18 mph and covers 1 mile

We want to know the total ave speed, we cant use Weighted averages : (12x2 + 18x1)/3 = 14 mph
because the 2 quaintities i.e. Speed units denominator (i.e. hours) does not match with Distance denominator (i.e. nothing).

Case 2.
Car moves A to B @ 12mph for 2 hrs
B to C @ 18 mph for 1 hr

We want to know the total ave speed, we can use Weighted averages : (12x2 + 18x1)/3 = 14 mph

But in case 2 the units denominators donot match, speed is mph & time is h? What I m missing here!

You are right. Case 1 doesn't work but case 2 does. What is the unit of the denominator in average speed? Average speed = miles/hour. So you can use weighted averages when the weights are time (i.e. in hours). That is the reason your case 2 works. (By the way, you yourself said speed is miles/hour and time is in hours. Why do you think the units don't match?)

When you want to find weighted average speed, distance cannot be the weights, it has to be time.
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Re: A certain car traveled twice as many miles from Town A  [#permalink]

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15 Feb 2015, 15:43
Hello Experts!
I am quiet fascinated with The Bunuel approach... of using algebraic expression.
Kindly assess my approach which I will try next time onward to got a shortcut way
step 1) took LCM of 12 and 18.. came as 36. just multiplied by 10...(to make easy calculation)
step 2) 360 distance between B to C... do 360/18 hence 20 gallons used
step 3) twice distance.. hence 360*2= 720... do as above.. 720/12= 60 gallons used
step 4) total gallons.. 20+60= 80 gallons
step ) total miles= 360+720= 1080 miles
hence.. average of whole journey = 1080/80 which comes to 13.5
Thanks
Celestial

Kudos if it makes sense...
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Re: A certain car traveled twice as many miles from Town A  [#permalink]

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15 Aug 2016, 00:57
Celestial09 wrote:
Hello Experts!
I am quiet fascinated with The Bunuel approach... of using algebraic expression.
Kindly assess my approach which I will try next time onward to got a shortcut way
step 1) took LCM of 12 and 18.. came as 36. just multiplied by 10...(to make easy calculation)
step 2) 360 distance between B to C... do 360/18 hence 20 gallons used
step 3) twice distance.. hence 360*2= 720... do as above.. 720/12= 60 gallons used
step 4) total gallons.. 20+60= 80 gallons
step ) total miles= 360+720= 1080 miles
hence.. average of whole journey = 1080/80 which comes to 13.5
Thanks
Celestial

Kudos if it makes sense...

While you approach is not wrong, you have made lots of double and tedious calculations which is quite time consuming. You can do it without multiplying by 10

You got LCM 36. so we will considered total miles traveled under 18 Miles/gallon to be 36. Hence it took 2 gallons
Now, the distance traveled under 12M/ Gallon will be 72 and total gallons used will be 6.

therefore no. of gallons used traveling both route = 2+6 -= 8 and
total miles traveled is 36+72 = 108.

average = 108/8= 13.5 M/Gallon

its much more easier than you steps, which include unnecessary multiplication and double division. Its quite time consuming in GMAT.
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Re: A certain car traveled twice as many miles from Town A  [#permalink]

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29 Oct 2017, 23:42
plugin a number for distance
Let B to c=10
then as per question A to B=20
Total distance=30

Fuel used
B to C=10/18=8/9(Distance/Mileage per Gallon)
A to B=20/12=5/3

Total mileage=Total distance/Total fuel used=30/(5/3)+(8/9)=27/9=13.5
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Re: A certain car traveled twice as many miles from Town A  [#permalink]

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31 Oct 2017, 16:15
docabuzar wrote:
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 18 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C?

A. 13
B. 13.5
C. 14
D. 14.5
E. 15

We can use the following formula:

Average = (total distance)/(total gallons)

We can let the distance from Town A to Town B = 2d and the distance from Town B to Town C = d. Thus, the total gallons of fuel consumed from Town A to Town B = 2d/12, and the total gallons of fuel consumed from Town B to Town C = d/18; thus:

Average = 3d/(2d/12 + d/18)

Average = 3d/(6d/36 + 2d/36)

Average = 3d/(8d/36) = (36 x 3d)/8d = (9 x 3)/2 = 13.5

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Re: A certain car traveled twice as many miles from Town A &nbs [#permalink] 31 Oct 2017, 16:15
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